Yes, you are totally correct.

your guard for length zero should be ([],[]). The type of the function is

halve :: [a] -> ([a], [a])

so all the code paths have to finish with that type.

Incidentally taking the length of a linked list forces you to walk the entire list. You are better off checking as to whether it is empty or not.

Regards,
Ben

On 12 July 2011 10:44, Roelof Wobben <rwobben@hotmail.com> wrote:

Oke,



I have now this as function definition.





halve (xs) | length xs `mod` 2 == 0   = (take n xs, drop n xs)
          | otherwise = (take (n+1) xs,  drop (n+1) xs)
 where n= length xs `div` 2


main = do
 putStrLn $ show $ halve [1,2,3,4]
 putStrLn $ show $ halve [1,2,3]




this one works except for empty lists.

So I thought this would work .



halve (xs) | length xs == 0 = []

          | length xs `mod`2 == 0 = (take n xs, drop n xs)

          | otherwise = (take (n+1) xs, drop (n+1) xs)

  where n = length xs `div`2





but then I see this error :





Error occurred
ERROR line 2 - Type error in guarded expression
*** Term           : (take n xs,drop n xs)
*** Type           : ([b],[b])
*** Does not match : [a]



So I assume that a function must always have the same output and can't have 1 or 2 lists as output.



Is this the right assumption.



Roelof


________________________________
> Date: Tue, 12 Jul 2011 10:34:25 +0100
> Subject: Re: [Haskell-beginners] function defenition. Do I understand
> it right?
> From: edwards.benj@gmail.com
> To: rwobben@hotmail.com
> CC: beginners@haskell.org
>
> I don't even understand what you are trying to do :)
>
> if you want to pattern match on the empty list
>
> foo :: [a] -> [a]
> foo [] = 0
> foo (x:xs) = undefined
>
> if you want to use the guard syntax
>
> foo xs | null xs = 0
> | otherwise = undefined
>
>
> Ben
>
> On 12 July 2011 10:02, Roelof Wobben
> <rwobben@hotmail.com<mailto:rwobben@hotmail.com>> wrote:
>
>
>
> hello
>
>
>
> Everyone thanks for the help.
>
> I'm now trying to make this work on a empty list.
>
>
>
> But my question is.
>
> When the definition is :
>
>
>
> [a] -> [a] [a]
>
>
>
> Is it correct that I don't can use.
>
>
>
> length xs = 0 | []
>
>
>
> Roelof
>
> ----------------------------------------
> > Subject: Re: [Haskell-beginners] function defenition. Do I understand
> it right?
> > From: d@vidplace.com<mailto:d@vidplace.com>
> > Date: Mon, 11 Jul 2011 18:56:42 -0400
> > CC: beginners@haskell.org<mailto:beginners@haskell.org>
> > To: rwobben@hotmail.com<mailto:rwobben@hotmail.com>
> >
> > On Jul 11, 2011, at 5:13 PM, Roelof Wobben wrote:
> >
> > > What I try to achieve is this:
> > >
> > >
> > >
> > > [1,2,3,4] will be [1,2] [3,4]
> > >
> > > [1,2,3,4,5] will be [1,2,3] [3,4,5]
> >
> > So, I think what you want is this http://codepad.org/kjpbtLfR
> >
> > Is that correct?
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org<mailto:Beginners@haskell.org>
> http://www.haskell.org/mailman/listinfo/beginners
>
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